MCE 511 Finite Element Analysis - OA...

MCE 511 Finite Element Analysis01 Variational Calculus

Instructor: OA [email protected]

Department of Systems EngineeringUniversity of Lagos

mailto:[email protected]

[email protected] 12/31/2012Department of Systems Engineering, University of Lagos 2

Scope of Instructional Material

Course Schedule:

The read-ahead materials are from Reddy except the part marked red. There please read Wolfram Manuals. Home work assignments will be drawn from the range of pages in the respective books.

Slide Title Slides Wks Text Read Pages

Variational Calculus Intro 70 1 Reddy 1-57

Integral Formulation 50 1 Reddy 58-102

1-D Models 50 1 Reddy 103-154

Applications 60 1 Reddy 155-231

Computer Applications 50 3 Wolfram 109-129

1. Methods of Mathematical Physics, R. Courant & D. Hilbert, Vol1, Wiley Interscience, 1975

2. An Introduction to the Finite Element Method, JN Reddy, Third Edition, McGraw-Hill Education, 2006

3. The Variational Principles of Mechanics, C Lanczos, Fourth Edition, Dover Publications Inc, 1970

4. Install a copy of Mathematica on your computer to prepare for the computational part of the course. The Help System has all the literature you need for the computational methods

Wednesday, September 6, [email protected], University of Lagos 3

Recommended Texts

Provide the basis of the Finite Element Method for beginning graduate Students

Show the formulation of Practical Mechanics Problems

Build Finite Element Solutions with a Symbolics/Numerical Processor

Introduction to Finite Element Simulations.

Purpose of the Lecture


Consider the following problem in Geodesics:


Variations & the Geodesic Problem

We are given two fixed points A and B on the surface, 𝑧 =𝜙(𝑥, 𝑦). We desire to find the shortest distance from A to B.

Projected to the x-y plane, particular continuous, single-valued paths may appear like the any of the following.

It is no longer imperative, for example, that the straight line is the shortest.

This is an example of a problem that can be solved by the Calculus of Variations


Shortest Distance

Minimize the integral,

𝐽 = න𝑎

𝑏

𝑑𝑠 = න𝑎

𝑏

𝑑𝑥2 + 𝑑𝑦2 + 𝑑𝑧2

= න𝑎

𝑏

1 +𝑑𝑦

𝑑𝑥

2

+𝑑𝑧

𝑑𝑥

2

𝑑𝑥

But we are given that, along the surface, 𝑧 = 𝜙(𝑥, 𝑦) . Therefore,

𝑑𝑧 =𝜕𝜙

𝜕𝑥𝑑𝑥 +

𝜕𝜙

𝜕𝑦𝑑𝑦

Let the image of the path on the 𝑥 − 𝑦 plane be 𝑦 = 𝜓(𝑥) so that

𝑑𝑦

𝑑𝑥= 𝜓′ 𝑥


Shortest Distance

We therefore find that

𝑑𝑧 =𝜕𝜙

𝜕𝑥𝑑𝑥 +

𝜕𝜙

𝜕𝑦𝑑𝑦

=𝜕𝜙

𝜕𝑥𝑑𝑥 +

𝜕𝜙

𝜕𝑦

𝑑𝑦

𝑑𝑥𝑑𝑥.

So that, we are to minimize the integral

𝐽 = න𝑎

𝑏

1 +𝑑𝜓

𝑑𝑥

2

+𝜕𝜙

𝜕𝑥+𝜕𝜙

𝜕𝑦

𝑑𝑦

𝑑𝑥

2

𝑑𝑥

= න𝑎

𝑏

1 + 𝜓′ 𝑥2+

𝜕𝜙 𝑥, 𝜓 𝑥

𝜕𝑥+𝜕𝜙 𝑥, 𝜓 𝑥

𝜕𝑦𝜓′ 𝑥

2

𝑑𝑥

= න𝑎

𝑏

𝐹 𝑥, 𝜓 𝑥 ,𝜓′ 𝑥 𝑑𝑥

by finding the best function, 𝑦 = 𝜓(𝑥), among the set of admissible functions. Recall that we already know the function 𝑧 = 𝜙 𝑥, 𝑦 .


Shortest Distance

There have been a large number of problems that can be formulated in a similar way. Variational Calculus has a rich history of problems that have engaged the most fertile mathematical geniuses over the years. Examples include:

Dido’s Problem, or the Isoperimetric problem,

The brachistochrone,

Minimum Surface,

Fermat’s principle of least time,

Lagrangian Principle of least action and other minimization principles in Mechanics.


Other Historical Problems

Enclose the largest possible area of land within a fence with a predetermined length. The problem becomes more interesting if we allow a weight function 𝜌 𝑥 𝑦 to be defined at each point. Such a function may represent things such as fertility or the concentration of some valuable quantity. Mathematically, this is equivalent to finding the closed curve of a given length that extremises the integral 𝜌 𝑥, 𝑦 𝑑𝑥𝑑𝑦.


Dido’s Problem

The Brachistochrone. A frictionless ring attached to a fixed smooth rod is falling under the influence of gravity from point A to point B. What shape of the rod will make the travel time smallest? We show later that this curve is a catenary.


The Brachistochrone

Minimal Surface. A plane curve joining given points A and B is to be rotated about an axis. We desire to know the particular curve out of all the possible curves that can be made to pass these two points, that will produce the smallest surface area.


Minimal Surface

From differential and integral Calculus, we are used the concept of functions. Given a domain 𝑎, 𝑏 , we can define a function that take

values such as 𝑦 = 𝑥2 in the domain.

We may use the methods of calculus to find the maxima or minima of such a function, if any exists, in the domain.

We can define multi-valued functions in the same way.

In each case, we are mapping from the real space or a product real space to the real space.

The geodesic problem we saw earlier has something fundamentally different. It is called a functional.


Functions and Functionals

In the minimization of

𝐽 = න𝑎

𝑏

𝐹 𝑥, 𝜓′ 𝑥 ,𝜕𝜙 𝑥, 𝜓 𝑥

𝜕𝑥,𝜕𝜙 𝑥, 𝜓 𝑥

𝜕𝑦𝑑𝑥

We are trying to find a function that minimizes the integral. Unlike our example in calculus, it is a function that will minimize our integral rather than a particular point! In this case, we are really looking for the function 𝑦 = 𝜓(𝑥) that minimizes the functional which is a function of the function 𝜓(𝑥) and some of its derivatives.

Variational Calculus deals with the extremal values of functionals.


What is a Functional?

The simplest problem of Calculus of Variations is to find the function, 𝑦 = 𝜓 𝑥 , that extremizes the Integral,

𝐽[𝑦 𝑥 ] = න𝑎

𝑏

𝐹 𝑥, 𝑦 𝑥 , 𝑦′ 𝑥 𝑑𝑥

𝑥 ∈ 𝑎, 𝑏 , 𝑦 𝑎 = 𝑦𝑎, and 𝑦 𝑏 = 𝑦𝑏

We talk of extremization because sometimes we are looking for minimum and sometimes we are looking for maximum.

Such extreme values always occur at the stationary points of the functional under examination.


The Simplest Problem

We assume that the functions that will qualify satisfy certain basic smoothness conditions, are single-valued.

Each function must necessarily satisfy, a priori, certain end conditions such that 𝑦 𝑎 = 𝑦𝑎 , and 𝑦 𝑏 = 𝑦𝑏 with the values of 𝑦𝑎, and 𝑦𝑏 supplied in advance.

Such explicit conditions are called “Essential Boundary Conditions” as they must be satisfied by any trial functions we shall choose.

Other boundary conditions will arise later. They will be called “Natural Boundary Conditions”. Suffice it is to say that any boundary conditions that cannot be classified as “Essential” will be in the second category.


Admissible functions

The following functions are NOT admissible: 1 and 3 are not smooth, 2 is discontinuous, 4 is not single-valued and five does not satisfy the essential boundary conditions.


Inadmissible Functions

In order to solve the problem, we consider the rate of change in the functional resulting in a variation in the admissible trial function 𝑦 = 𝑦(𝑥).

We shall be specific and ensure that each varied path satisfies the admissibility conditions as stated earlier.

In particular, assume that 𝑦 = 𝑦(𝑥) is admissible, Then introduce the function 𝜙 𝑥 which is such that 𝜙 𝑎 = 𝜙 𝑏 = 0, then, it follows that, the function

𝑦 𝑥 + 𝜖𝜙 𝑥 ≡ 𝑦 + 𝛿𝑦


The First Variation

𝑦 𝑥 + 𝜖𝜙 𝑥 ≡ 𝑦 + 𝛿𝑦

Is also admissible provided 𝜙 𝑥 also satisfies the smoothness conditions stated earlier. Moreover, we can make 𝜖 as small as we desire and in the limit, as 𝜖 →0, this function approaches 𝑦(𝑥).


The First Variation

The variation in the functional as a result of variation in the paths from the extremizing path is,

𝛿𝐹 𝑥, 𝑦, 𝑦′ ≡ 𝐹 𝑥, 𝑦 + 𝛿𝑦, 𝑦′ + 𝛿𝑦′ − 𝐹 𝑥, 𝑦, 𝑦′

Note that there is no variation in the independent variable; the above variations are only in 𝑦 and 𝑦′. We can therefore write,

𝛿𝐹 𝑥, 𝑦, 𝑦′ =𝜕𝐹

𝜕𝑦𝛿𝑦 +

𝜕𝐹

𝜕𝑦′𝛿𝑦′ = 𝜖

𝜕𝐹

𝜕𝑦𝜙 +

𝜕𝐹

𝜕𝑦′𝜙′


The First Variation

Where we have only taken the first two terms of the Taylor Series,

𝐹 𝑥 + Δ𝑥, 𝑦 + Δ𝑦

= 𝐹 𝑥 + Δ𝑥, 𝑦 + Δ𝑦 +𝜕𝐹

𝜕𝑥Δ𝑥 +

𝜕𝐹

𝜕𝑦Δ𝑦

+1

2!

𝜕2𝐹

𝜕𝑥2Δ𝑥 2 +

2𝜕2𝐹

𝜕𝑥𝜕𝑦Δ𝑥Δ𝑦 +

𝜕2𝐹

𝜕𝑦2𝛥𝑦 2 +⋯


Taylor Series

Upon Integrating by parts, we can write,

𝛿𝐽

𝜖= න

𝑎

𝑏 𝜕𝐹

𝜕𝑦𝜙 +

𝜕𝐹

𝜕𝑦′𝜙′ 𝑑𝑥

= න𝑎

𝑏 𝜕𝐹

𝜕𝑦−

𝑑

𝑑𝑥

𝜕𝐹

𝜕𝑦′𝜙𝑑𝑥 +

𝜕𝐹

𝜕𝑦′𝜙

𝑎

𝑏

Coming from the fact that,

න𝑎

𝑏 𝜕𝐹

𝜕𝑦′𝜙′ 𝑑𝑥 =

𝜕𝐹

𝜕𝑦′𝜙

𝑎

𝑏

−න𝑎

𝑏 𝑑

𝑑𝑥

𝜕𝐹

𝜕𝑦′𝜙𝑑𝑥


First Variation

We arrived at the expression of this variation by ignoring higher terms than the second in the Taylor Series. It is the first variation that must vanish in order for an extremum to exist at 𝑦 = 𝑦(𝑥). It can be shown, by an invocation of the fundamental theorem of integral calculus that this implies that,

𝜕𝐹

𝜕𝑦−

𝑑

𝑑𝑥

𝜕𝐹

𝜕𝑦′= 0

The essential boundary conditions, that 𝜙 𝑎 = 𝜙 𝑏 =

0, ensures that 𝜕𝐹

𝜕𝑦′𝜙

𝑎

𝑏= 0 automatically.


Euler-Lagrange Equations

The divergence theorem is central to several other results in Continuum Mechanics. We present here a generalized form [Ogden] which states that,

Gauss Divergence Theorem

For a tensor field 𝚵, The volume integral in the region Ω ⊂ ℰ,

නΩ

grad 𝚵 𝑑𝑣 = න𝜕Ω

𝚵⊗ 𝐧𝑑𝑎

where 𝐧 is the outward drawn normal to 𝜕Ω – the boundary of Ω.

Integral Theorems

[email protected] 12/27/2012Dept of Systems Engineering, University of Lagos 24

Consider a cylindrical volumetric domain such that the volume element lies between two surfaces of a cylinder as shown. It is assumed that the depth of the cylinder 𝑏 → 0. We can therefore write, 𝑑𝑣 =𝑏𝑑𝐴, and 𝑑𝑎 = 𝑏𝑑𝑠Clearly,

න𝜕C

grad 𝜩 𝑑𝐴 = ර𝜩⊗ 𝒏 𝑑𝑠

Where we have assumed that the area integral on the two lateral surfaces will cancel out as 𝑏 → 0 and the outwardly drawn normals oppose.


2-D Version

Vector field. Replacing the tensor with the vector field 𝐟and contracting, we have,

නΩ

div 𝐟 𝑑𝑣 = න𝜕Ω

𝐟 ⋅ 𝐧 𝑑𝑎

Which is the usual form of the Gauss theorem. In 2-D, this becomes,

නΩ

div 𝐟 𝑑𝐴 = ර𝐟 ⋅ 𝐧 𝑑𝑠

Special Cases: Vector Field


For a scalar field 𝜙, the divergence becomes a gradient and the scalar product on the RHS becomes a simple multiplication. Hence the divergence theorem becomes,

නΩ

grad 𝜙 𝑑𝑣 = න𝜕Ω

𝜙𝐧𝑑𝑎

The procedure here is valid and will become obvious if we write, 𝐟 = 𝜙𝒃 where 𝒃 is an arbitrary constantvector.

Scalar Field


නΩ

div 𝜙𝒃 𝑑𝑣 = න𝜕Ω

𝜙𝒃 ⋅ 𝐧 𝑑𝑎 = 𝒃 ⋅ න𝜕Ω

𝜙𝐧𝑑𝑎

For the LHS, note that, div 𝜙𝒃 = tr grad 𝜙𝒃

grad 𝜙𝒃 = 𝜙𝑏𝑖 ,𝑗 𝐠𝑖 ⊗𝐠𝑗

= 𝑏𝑖𝜙,𝑗 𝐠𝑖 ⊗𝐠𝑗

The trace of which is,

𝑏𝑖𝜙,𝑗 𝐠𝑖 ⋅ 𝐠𝑗 = 𝑏𝑖𝜙,𝑗 𝛿𝑖

𝑗

= 𝑏𝑖𝜙,𝑖 = 𝒃 ⋅ grad 𝜙

For the arbitrary constant vector 𝒃, we therefore have that,

නΩ

div 𝑑𝑣 = 𝒃 ⋅ නΩ

grad 𝜙 𝑑𝑣 = 𝒃 ⋅ න𝜕Ω

𝜙𝐧𝑑𝑎

Ω grad 𝜙 𝑑𝑣 = Ω𝜙𝐧𝑑𝑎�� or Ω grad 𝜙 𝑑𝐴 = 𝜙𝒏ׯ 𝑑𝑠


නΩ

grad 𝜙 𝑑𝐴 = ර𝜙𝒏 𝑑𝑠

From here we can write that,

නΩ

grad 𝜙𝜓 𝑑𝐴 = නΩ

𝜙 grad 𝜓𝑑𝐴 +නΩ

𝜓 grad 𝜙 𝑑𝐴 = ර𝜙𝜓 𝒏 𝑑𝑠

So that,

නΩ

𝜓 grad 𝜙𝑑𝐴 = ර𝜙𝜓 𝒏 𝑑𝑠 − නΩ

𝜙 grad 𝜓𝑑𝐴.

We can break this, in Cartesian coordinates, to its two components:

නΩ

𝜓𝜕𝜙

𝜕𝑥𝑑𝐴 = ර𝜙𝜓 𝑛𝑥𝑑𝑠 − න

Ω

𝜙𝜕𝜓

𝜕𝑥𝑑𝐴

නΩ

𝜓𝜕𝜙

𝜕𝑦𝑑𝐴 = ර𝜙𝜓 𝑛𝑦𝑑𝑠 − න

Ω

𝜙𝜕𝜓

𝜕𝑦𝑑𝐴


Further Results (Scalar)

Beginning from the standard Gauss theorem in 2-D,

නΩ

div 𝐟 𝑑𝐴 = ර𝐟 ⋅ 𝐧 𝑑𝑠

Now, let 𝐟 = 𝑤𝐅 where 𝐅 is a differentiable vector field and 𝑤 is a scalar field. Then we can write,

නΩ

div 𝑤𝐅 𝑑𝐴 = ර𝑤𝐅 ⋅ 𝐧 𝑑𝑠

But div 𝑤𝐅 = grad 𝑤 ⋅ 𝐅 + 𝑤 div 𝐅. If we are also given that the field 𝐅 is derived from a scalar potential 𝜙, so that 𝐅 = grad 𝜙, then we can write that,

නΩ

grad 𝑤 ⋅ grad 𝜙 𝑑𝐴 + නΩ

𝑤 div grad 𝜙 𝑑𝐴 = ර𝑤 grad 𝜙 ⋅ 𝐧 𝑑𝑠

Which can be rearranged to obtain,

−නΩ

𝑤 grad2 𝜙𝑑𝐴 = නΩ

grad 𝑤 ⋅ grad 𝜙 𝑑𝐴 − ර𝑤 grad 𝜙 ⋅ 𝐧 𝑑𝑠


Further Results (Vector)

For a second-order tensor 𝐓, the Gauss Theorem becomes,

නΩ

div𝐓 𝑑𝑣 = න𝜕Ω

𝐓𝐧𝑑𝑎

The original outer product under the integral can be expressed in dyadic form:

නΩ

grad 𝐓 𝑑𝑣 = නΩ

𝑇𝑖𝑗 ,𝑘 𝐠𝑖 ⊗𝐠𝑗 ⊗𝐠𝑘 𝑑𝑣

= න𝜕Ω

𝐓⊗ 𝐧𝑑𝑎

= න𝜕Ω

𝑇𝑖𝑗𝐠𝑖 ⊗𝐠𝑗 ⊗ 𝑛𝑘𝐠𝑘 𝑑𝑎

Second-Order Tensor field


Or

නΩ

𝑇𝑖𝑗 ,𝑘 𝐠𝑖 ⊗𝐠𝑗 ⊗𝐠𝑘𝑑𝑣 = න𝜕Ω

𝑇𝑖𝑗𝑛𝑘𝐠𝑖 ⊗𝐠𝑗 ⊗𝐠𝑘 𝑑𝑎

Contracting, we have

නΩ

𝑇𝑖𝑗 ,𝑘 𝐠𝑖 ⊗𝐠𝑗 𝐠𝑘𝑑𝑣 = න𝜕Ω

𝑇𝑖𝑗𝑛𝑘 𝐠𝑖 ⊗𝐠𝑗 𝐠𝑘 𝑑𝑎

නΩ

𝑇𝑖𝑗 ,𝑘 𝛿𝑗𝑘𝐠𝑖 𝑑𝑣 = න

𝜕Ω

𝑇𝑖𝑗𝑛𝑘𝛿𝑗𝑘𝐠𝑖 𝑑𝑎

නΩ

𝑇𝑖𝑗 ,𝑗 𝐠𝑖 𝑑𝑣 = න𝜕Ω

𝑇𝑖𝑗𝑛𝑗𝐠𝑖 𝑑𝑎

Which is the same as,

නΩ

div𝐓 𝑑𝑣 = න𝜕Ω

𝐓𝐧𝑑𝑎

Second-Order Tensor field


Consider the Euclidean Point Space ℰ. A curve 𝒞 is defined by the parametrization (Gurtin):

𝐱 = ො𝐱 𝜆 where 𝜆0 ≤ 𝜆 ∈ ℛ ≤ 𝜆1𝒞 is said to be a closed curve if ො𝐱 𝜆0 = ො𝐱 𝜆1

Define 𝐭 𝜆 ≡𝑑ො𝐱 𝜆

𝑑𝜆.

For any vector point function defined everywhere along 𝒞, the line integral,

න𝒞

𝐯 ⋅ 𝒅𝐱 = න𝜆0

𝜆1

𝐯 ො𝐱 𝜆0 ⋅𝑑ො𝐱 𝜆

𝑑𝜆𝑑𝜆 =න

𝜆0

𝜆1

𝐯 ො𝐱 𝜆0 ⋅ 𝐭 𝜆 𝑑𝜆


Stokes Theorem


Stokes Theorem

Let 𝜙 be a scalar field on ℰ, the

Chain rule immediately implies

that,

න𝜆0

𝜆1

grad 𝜙 ⋅ 𝒅𝐱 = න𝜆0

𝜆1

grad 𝜙 ⋅𝑑ො𝐱 𝜆

𝑑𝜆𝑑𝜆

= න𝜆0

𝜆1 𝜕𝜙 ො𝐱 𝜆

𝜕𝜆𝑑𝜆 = 𝜙 ො𝐱 𝜆1 − 𝜙 ො𝐱 𝜆0

So that for a close curve 𝒞 grad 𝜙 ⋅ 𝒅𝐱 = 0

For a positively oriented surface bounded by a closed curve 𝒞

(Gurtin), Stokes theorem is stated as follows:


Stokes Theorem

Stokes’ Theorem Let 𝜙, 𝐯, and 𝐓 be scalar, vector, and tensor fields with common domain a region ℛ. Then given any positively oriented surface 𝒮, with boundary 𝒞 closed curve, in ℛ

න𝒞

𝜙𝑑𝐱 =න𝒮

𝐧 × grad 𝜙 𝑑𝑎

න𝒞

𝐯 ⋅ 𝑑𝐱 =න𝒮

𝐧 × curl 𝐯 𝑑𝑎

න𝒞

𝐓𝑑𝐱 =න𝒮

curl 𝐓 T𝑑𝑎

𝜕𝐹

𝜕𝑦−

𝑑

𝑑𝑥

𝜕𝐹

𝜕𝑦′= 0

Noting the fact that,𝜕𝐹

𝜕𝑦′= Ψ 𝑥, 𝑦 𝑥 , 𝑦′(𝑥) ,

Explicitly showing it is a function of the three variables. The total derivative of

𝑑Ψ =𝜕Ψ

𝜕𝑥𝑑𝑥 +

𝜕Ψ

𝜕𝑦𝑑𝑦 +

𝜕Ψ

𝜕𝑦′𝑑𝑦′

So that,𝑑

𝑑𝑥

𝜕𝐹

𝜕𝑦′=𝜕Ψ

𝜕𝑥+𝜕Ψ

𝜕𝑦𝑦′ +

𝜕Ψ

𝜕𝑦′𝑦′′ =

𝜕2𝐹

𝜕𝑥𝜕𝑦′+

𝜕2𝐹

𝜕𝑦𝜕𝑦′𝑦′ +

𝜕2𝐹

𝜕𝑦′𝜕𝑦′𝑦′′


Simplest Problem Explicit

So that the minimization condition becomes,𝜕2𝐹

𝜕𝑥𝜕𝑦′+

𝜕2𝐹

𝜕𝑦𝜕𝑦′𝑦′ +

𝜕2𝐹

𝜕𝑦′𝜕𝑦′𝑦′′ −

𝜕𝐹

𝜕𝑦= 0.

This expression is also called the “variational derivative” of the functional:

𝐹 𝑦 𝑥 =𝜕𝐹

𝜕𝑦−

𝑑

𝑑𝑥

𝜕𝐹

𝜕𝑦′

= −𝜕2𝐹

𝜕𝑥𝜕𝑦′−

𝜕2𝐹

𝜕𝑦𝜕𝑦′𝑦′ −

𝜕2𝐹

𝜕𝑦′𝜕𝑦′𝑦′′ +

𝜕𝐹

𝜕𝑦= 0

For example, in the plane geodesy problem, only 𝑦′ appears explicitly in the functional. Hence, the extremizing condition is simply,

𝜕2𝐹

𝜕𝑦′𝜕𝑦′𝑦′′ = 0

From which we easily recover the obvious fact that only a straight line can minimize the distance on a flat plane.


Simplest Problem

Recall that the plane geodesic problem is simply the extremization of the integral,

𝐽 = 1 +𝑑𝑦

𝑑𝑥

2

𝑑𝑥

Where the functional 𝐹 = 𝐹 𝑦′(𝑥) which is the simplest problem with only the derivative of the independent function appearing in the functional. Hence, all other terms in the variational derivative vanishes and the extremizing condition is simply,

𝜕2𝐹

𝜕𝑦′𝜕𝑦′

𝑑2𝑦

𝑑𝑥2= 0

From which the well-known fact that a straight line is the shortest distance between two points on a flat plane immediately emerges.


Plane Geodesic

The foregoing can be extended to include the case of functionals with two or more functional arguments. Consider the case where the function

𝐽 𝑦1 𝑥 ,… , 𝑦𝑛(𝑥)

= න𝑥0

𝑥1

𝐹 𝑥, 𝑦1 𝑥 ,… , 𝑦𝑛(𝑥), ሶ𝑦1 𝑥 ,… , ሶ𝑦𝑛(𝑥) 𝑑𝑥

Where we have used dots to denote derivatives with respect to the independent variable 𝑥 so that the dependent variable arguments of the functional are 𝑦1 𝑥 ,… , 𝑦𝑛(𝑥) and their first derivatives ሶ𝑦1 𝑥 ,… , ሶ𝑦𝑛(𝑥). We assume that the continuity requirements for the single function case remain valid in this case for all the independent and function arguments shown. We now suppose that functions


Several Independent Functions


𝑦𝑖 𝑥 = 𝑓𝑖 𝑥 , 𝑖 = 1,2, …𝑛exist in the interval 𝑥0 ≤ 𝑥 ≤ 𝑥1with prescribed values at the endpoints of this interval which are such that the above functional assumes an extreme value in comparison with other admissible sets of functions in a predetermined neighbourhood of the set of

functions in 𝑓𝑖 𝑥 , 𝑖 = 1,2, …𝑛 satisfying the same prescribed conditions at the endpoints. For this to happen, the first variation of the definite integral must vanish. This immediately leads to the set of equations,

𝛿𝐽 = න𝑥0

𝑥1 𝜕𝐹

𝜕𝑦𝑖𝛿𝑦𝑖 +

𝜕𝐹

𝜕 ሶ𝑦𝑖𝛿 ሶ𝑦𝑖 𝑑𝑥 = 0


Note that each term in the brackets is actually a sum of 𝑛 terms by the Einstein summation convention. Integrating the second sum by parts, we have

න𝑥0

𝑥1 𝜕𝐹

𝜕 ሶ𝑦𝑖𝛿 ሶ𝑦𝑖 𝑑𝑥 =

𝜕𝐹

𝜕 ሶ𝑦𝑖𝛿𝑦𝑖

𝑥0

𝑥1

−න𝑥0

𝑥1 𝑑

𝑑𝑥

𝜕𝐹

𝜕 ሶ𝑦𝑖𝛿𝑦𝑖 𝑑𝑥

The first term vanishes as usual on account of the vanishing of the variations in the dependent variables at the endpoints. Substituting this back into the minimization condition, we can write,

𝛿𝐽 = න𝑥0

𝑥1 𝜕𝐹

𝜕𝑦𝑖−

𝑑

𝑑𝑥

𝜕𝐹

𝜕 ሶ𝑦𝑖𝛿𝑦𝑖𝑑𝑥 = 0

But the variations in the functional arguments are all applied independently.


The fundamental lemma of variational calculus therefore requires the vanishing of the terms in the square brackets for each 𝑖 =1, 2, . . . , 𝑛 so that each variational derivative vanishes:

𝜕𝐹

𝜕𝑦𝑖−

𝜕

𝜕𝑥

𝜕𝐹

𝜕 ሶ𝑦𝑖= 0

The functional 𝐹 here is a function of 2𝑛 + 1 variables 𝑦1 𝑥 ,… , 𝑦𝑛(𝑥) and their first derivatives ሶ𝑦1 𝑥 ,… , ሶ𝑦𝑛(𝑥). In analytical mechanics, if the independent variable 𝑥 is the elapsed

time 𝑡, 𝐹 represents the work function, the 𝑦𝑖 𝑥 𝑠 ’s are the generalized coordinates and the ሶ𝑦𝑖 𝑥 𝑠 are the generalized velocities. These equations are then are the celebrated Euler-Lagrange equations of motion.

Consider the case where the functional has another term of the second order:

𝐽 𝑦 𝑥 = න𝑥0

𝑥1

𝐹 𝑥, 𝑦 𝑥 , 𝑦′(𝑥), 𝑦′′(𝑥) 𝑑𝑥.

A simple extension of the previous arguments with similar conditions lead us to the extremization condition,

𝛿𝐽 = න𝑥0

𝑥1 𝜕𝐹

𝜕𝑦𝛿𝑦 +

𝜕𝐹

𝜕𝑦′𝛿𝑦′ +

𝜕𝐹

𝜕𝑦′′𝛿𝑦′′ 𝑑𝑥 = 0

Integrating the second term by parts, we have,

න𝑥0

𝑥1 𝜕𝐹

𝜕𝑦′𝛿𝑦′ 𝑑𝑥 =

𝜕𝐹

𝜕𝑦′𝛿𝑦

𝑥0

𝑥1

−න𝑥0

𝑥1 𝑑

𝑑𝑥

𝜕𝐹

𝜕𝑦′𝛿𝑦𝑑𝑥


Higher-Order Derivatives


For the third term, we have,

න𝑥0

𝑥1 𝜕𝐹

𝜕𝑦′′𝛿𝑦′′ 𝑑𝑥 =

𝜕𝐹

𝜕𝑦′′𝛿𝑦′

𝑥0

𝑥1

−න𝑥0

𝑥1 𝑑

𝑑𝑥

𝜕𝐹

𝜕𝑦′′𝛿𝑦′𝑑𝑥

=𝜕𝐹

𝜕𝑦′′𝛿𝑦′

𝑥0

𝑥1

−𝑑

𝑑𝑥

𝜕𝐹

𝜕𝑦′′𝛿𝑦

𝑥0

𝑥1

Substituting back into the extremizing condition,

𝛿𝐽 = න𝑥0

𝑥1 𝜕𝐹

𝜕𝑦−

𝑑

𝑑𝑥

𝜕𝐹

𝜕𝑦′+

𝑑2

𝑑𝑥2𝜕𝐹

𝜕𝑦′′𝛿𝑦𝑑𝑥 +

𝜕𝐹

𝜕𝑦′𝛿𝑦 +

𝜕𝐹

𝜕𝑦′′𝛿𝑦′ −

𝑑

𝑑𝑥

𝜕𝐹


𝑥0

𝑥1

= 0


Substituting back into the extremizing condition,𝛿𝐽

= න𝑥0

𝑥1 𝜕𝐹

𝜕𝑦−

𝑑

𝑑𝑥

𝜕𝐹

𝜕𝑦′+

𝑑2

𝑑𝑥2𝜕𝐹

𝜕𝑦′′𝛿𝑦𝑑𝑥 +

𝜕𝐹


𝜕𝐹

𝜕𝑦′′𝛿𝑦′ −

𝑑

𝑑𝑥

𝜕𝐹


𝑥0

𝑥1

= 0Which contains the appropriate Euler’s equations

𝛿𝐽 = න𝑥0

𝑥1 𝜕𝐹

𝜕𝑦−

𝑑

𝑑𝑥

𝜕𝐹

𝜕𝑦′+

𝑑2

𝑑𝑥2𝜕𝐹

𝜕𝑦′′𝛿𝑦𝑑𝑥 = 0

And the boundary conditions,

𝜕𝐹


𝜕𝐹

𝜕𝑦′′𝛿𝑦′ −

𝑑

𝑑𝑥

𝜕𝐹


𝑥0

𝑥1

that normally vanish identically when there is no variation at the endpoints –that is, when the end conditions are fixed.


Several Independent Variables

The case of two variables 𝑥 and 𝑦 sufficiently illustrates what happens here:

𝐽 𝑢 𝑥 = ඵΩ

𝐹 𝑥, 𝑦, 𝑢 𝑥, 𝑦 , 𝑢𝑥 𝑥, 𝑦 , 𝑢𝑦(𝑥, 𝑦) 𝑑𝑥𝑑𝑦

Without any further ado, there will be no variations in the independent variables 𝑥 and 𝑦. The first variation immediately leads to the vanishing of the integral,

ඵΩ

𝜕𝐹

𝜕𝑢𝛿𝑢 +

𝜕𝐹

𝜕𝑢𝑥𝛿𝑢𝑥 +

𝜕𝐹

𝜕𝑢𝑦𝛿𝑢𝑦 𝑑𝑥𝑑𝑦 = 0


We change the order of integration in the extremizing condition, and obtain,

ඵΩ

𝜕𝐹

𝜕𝑢𝛿𝑢 +

𝜕𝐹

𝜕𝑢𝑥

𝜕𝛿𝑢

𝜕𝑥+

𝜕𝐹

𝜕𝑢𝑦

𝜕𝛿𝑢

𝜕𝑦𝑑𝑥𝑑𝑦 = 0

We now appeal to the Gauss Divergence Theorem (Slide 29) to obtain the necessary integration by parts here; recall,

නΩ

𝜓𝜕𝜙

𝜕𝑥𝑑𝐴 = ර𝜙𝜓 𝑛𝑥𝑑𝑠 − න

Ω

𝜙𝜕𝜓

𝜕𝑥𝑑𝐴

In the second term, substituting 𝜓 for 𝜕𝐹

𝜕𝑢𝑥and 𝜙 for 𝛿𝑢, it follows that,

ඵΩ

𝜕𝐹

𝜕𝑢𝑥

𝜕𝛿𝑢

𝜕𝑥𝑑𝑥𝑑𝑦 =ර

𝜕𝐹

𝜕𝑢𝑥𝑛𝑥𝛿𝑢𝑑𝑠 −ඵ

Ω

𝜕

𝜕𝑥

𝜕𝐹

𝜕𝑢𝑥𝛿𝑢𝑑𝑥𝑑𝑦

Applying the second part of the Gauss theorem to the third term leads to,

ඵΩ

𝜕𝐹

𝜕𝑢𝑦

𝜕𝛿𝑢

𝜕𝑦𝑑𝑥𝑑𝑦 =ර

𝜕𝐹

𝜕𝑢𝑦𝑛𝑦𝛿𝑢𝑑𝑠 −ඵ

Ω

𝜕

𝜕𝑦

𝜕𝐹

𝜕𝑢𝑦𝛿𝑢𝑑𝑥𝑑𝑦


Finally, we can write that,

𝛿𝐽 = ඵΩ

𝜕𝐹

𝜕𝑢−

𝜕

𝜕𝑥

𝜕𝐹

𝜕𝑢𝑥−

𝜕

𝜕𝑦

𝜕𝐹

𝜕𝑢𝑦𝛿𝑢𝑑𝑥𝑑𝑦 + ර

Γ

𝜕𝐹

𝜕𝑢𝑥𝑛𝑥 +

𝜕𝐹

𝜕𝑢𝑦𝑛𝑦 𝛿𝑢𝑑𝑠

(Notational inconsistency here: the subscript on 𝑛 is the component. On 𝑢, it is partial derivative wrt subscript)Let the green line denote the boundary andConsider an outward drawn normal 𝐧 = cos 𝛼 𝐢 + sin 𝛼 𝐣 in the figure below:

cos 𝛼 =𝑑𝑦

𝑑𝑠, sin 𝛼 = −

𝑑𝑥

𝑑𝑠so that the above boundary becomes,

රΓ

𝜕𝐹


𝜕𝐹

𝜕𝑢𝑦𝑛𝑦 𝛿𝑢𝑑𝑠 = ර

Γ

𝜕𝐹

𝜕𝑢𝑥

𝑑𝑦

𝑑𝑠−

𝜕𝐹

𝜕𝑢𝑦

𝑑𝑥

𝑑𝑠𝛿𝑢𝑑𝑠 =

= රΓ

𝜕𝐹

𝜕𝑢𝑥𝑑𝑦 −

𝜕𝐹

𝜕𝑢𝑦𝑑𝑥 𝛿𝑢 = 0


so that the above boundary becomes,

රΓ

𝜕𝐹


𝜕𝐹

𝜕𝑢𝑦𝑛𝑦 𝛿𝑢𝑑𝑠 = ර

Γ

𝜕𝐹

𝜕𝑢𝑥

𝑑𝑦

𝑑𝑠−

𝜕𝐹

𝜕𝑢𝑦

𝑑𝑥

𝑑𝑠𝛿𝑢𝑑𝑠

= Γׯ𝜕𝐹


𝜕𝐹

𝜕𝑢𝑦𝑑𝑥 𝛿𝑢 = 0

Finally, we can write that, 𝛿𝐽

= ඵΩ

𝜕𝐹

𝜕𝑢−

𝜕

𝜕𝑥

𝜕𝐹

𝜕𝑢𝑥−

𝜕

𝜕𝑦

𝜕𝐹

𝜕𝑢𝑦𝛿𝑢𝑑𝑥𝑑𝑦 + ර

Γ

𝜕𝐹


𝜕𝐹

𝜕𝑢𝑦𝑑𝑥 𝛿𝑢

= 0

Following the same logic, a functional with two multivariable functions

𝐽 𝑢 𝑥, 𝑦 , 𝑣(𝑥, 𝑦) = ඵ 𝐹 𝑥, 𝑦, 𝑢 𝑥, 𝑦 , 𝑣 𝑥, 𝑦 , 𝑢𝑥, 𝑢𝑦 , 𝑣𝑥, 𝑣𝑦) 𝑑𝑥𝑑𝑦.

With the primary variables 𝑢 𝑥, 𝑦 , 𝑣(𝑥, 𝑦) will lead to the equation,

𝛿𝐽 = ඵΩ

𝜕𝐹

𝜕𝑢−

𝜕

𝜕𝑥

𝜕𝐹

𝜕𝑢𝑥−

𝜕

𝜕𝑦

𝜕𝐹

𝜕𝑢𝑦𝛿𝑢 +

𝜕𝐹

𝜕𝑣−

𝜕

𝜕𝑥

𝜕𝐹

𝜕𝑣𝑥−

𝜕

𝜕𝑦

𝜕𝐹

𝜕𝑣𝑦𝛿𝑣 𝑑𝑥𝑑𝑦

+රΓ

𝜕𝐹


𝜕𝐹

𝜕𝑢𝑦𝑛𝑦 𝛿𝑢 +

𝜕𝐹

𝜕𝑣𝑥𝑛𝑥 +

𝜕𝐹

𝜕𝑣𝑦𝑛𝑦 𝛿𝑣 𝑑𝑠


Multi-Dimensional Systems


And the fact that the variations in the two functions 𝑢 𝑥, 𝑦 , 𝑣(𝑥, 𝑦) are applied independently implies that the coefficients of 𝛿𝑢 and 𝛿𝑣 must vanish:

𝜕𝐹

𝜕𝑢−

𝜕

𝜕𝑥

𝜕𝐹

𝜕𝑢𝑥−

𝜕

𝜕𝑦

𝜕𝐹

𝜕𝑢𝑦= 0

And 𝜕𝐹

𝜕𝑣−

𝜕

𝜕𝑥

𝜕𝐹

𝜕𝑣𝑥−

𝜕

𝜕𝑦

𝜕𝐹

𝜕𝑣𝑦= 0

And the boundary term,

රΓ

𝜕𝐹


𝜕𝐹


𝜕𝐹


𝜕𝐹


Must also vanish.

Integration by parts, achieved here by using the 2-D version of the Gauss Divergence theorem achieved two purposes:

1. Creates a “weak” form of the problem. Weak in the sense that the level of differentiation is transferred from one variable (weakening the smoothness requirements) to another.

2. Exposing another set of variables apart from the primary variables of interest.


Primary & Secondary Variables

In the original problem, the variables are the functions 𝑢 𝑥, 𝑦 𝑎𝑛𝑑 𝑣(𝑥, 𝑦). As a result of the Integration by parts, we now have new functions 𝑄 𝑥, 𝑦 𝑎𝑛𝑑 𝑄 𝑥, 𝑦 defined by the integrals,

𝑄 𝑥, 𝑦 ≡𝜕𝐹


𝜕𝐹

𝜕𝑢𝑦𝑛𝑦

and

R 𝑥, 𝑦 ≡𝜕𝐹


𝜕𝐹

𝜕𝑣𝑦𝑛𝑦

The prescription of the primary variables on the boundary are the Dirichlet, rigid, kinematic or Essential Boundary Conditions; the specification of the secondary variables are the Newman or Natural Boundary Conditions.


Primary & Secondary Variables

We now rewrite the variational extremization in terms of primary and secondary variables as follows:

1. First observe that in the interior, the primary variables must vanish; this leads naturally to the set of Euler-Lagrange equations:

𝜕𝐹

𝜕𝑢−

𝜕

𝜕𝑥

𝜕𝐹

𝜕𝑢𝑥−

𝜕

𝜕𝑦

𝜕𝐹

𝜕𝑢𝑦= 0 and

𝜕𝐹

𝜕𝑣−

𝜕

𝜕𝑥

𝜕𝐹

𝜕𝑣𝑥−

𝜕

𝜕𝑦

𝜕𝐹

𝜕𝑣𝑦= 0

2. And the boundary term,

රΓ

𝜕𝐹


𝜕𝐹


𝜕𝐹


𝜕𝐹


Now looks clearer in terms of the primary and secondary variables. If the primary variables are not specified, their variations will be non-zero. In orderfor the extremization to remain valid, their coefficients, the secondary variables will automatically vanish. On the boundary, the vanishing coefficients of variation of the primary variables (which are called secondary variables) give the natural boundary conditions.


One of the big confusions in the application of variational, and consequently, finite element, methods is in the practitioners’ inability to correctly articulate the issues of Boundary conditions.

In early elucidations on this issue, (See For example Courant & Hilbert I (pg 208-214), the matter, for the patient reader, is well spelt out.

We use the (Lanczos’) demonstration of a transversely-loaded beam to concretize these ideas.


Boundary Conditions

Instead of finding the optimal path from A to B, imagine we are now finding the best path from one end to the other with no restrictions.


The Extremized and varied paths may now look like the following:


Unrestricted Varied Paths

The end values are no longer prescribed. The varied paths still satisfy previous conditions for extremisation.

We consider a concrete example of this first before we present the theoretical treatment of the boundary conditions and its practical implications. Ref: Lanczos (68-73), Courant & Hilbert (208-214)

Consider the loaded beam as shown in different support configurations:


Concrete Example


The bending elastica of the above beam can be found by solving the ODE: Finding 𝑦(𝑥) that satisfies,

𝑑4𝑦

𝑑𝑥4=𝑤(𝑥)

𝑘where 𝑘 is a constant depending on the elastic property of the material and geometry. We derive this same equation from a variational principle:

The potential energy, 𝑉1 due to elastic forces and the that due to gravitational forces 𝑉2 are given by,

𝑉1 =𝑘

2න𝑥0

𝑥1

𝑦′′ 𝑥2𝑑𝑥 and 𝑉2 = −න

𝑥0

𝑥1

𝑤 𝑥 𝑦(𝑥)𝑑𝑥

The functional to minimize in this case is therefore,

𝐽 𝑦 𝑥 = න𝑥0

𝑥1 𝑘

2𝑦′′ 𝑥

2− 𝑤 𝑥 𝑦 𝑥 𝑑𝑥

= න𝑥0

𝑥1

𝐹(𝑥, 𝑦 𝑥 , 𝑦′′𝑥 )𝑑𝑥

The functional depending explicitly only on 𝑦(𝑥) and 𝑦′′ 𝑥 .


The variational derivative (Slide 41) of this functional is,

𝜕𝐹

𝜕𝑦−

𝑑

𝑑𝑥

𝜕𝐹

𝜕𝑦′+

𝑑2

𝑑𝑥2𝜕𝐹

𝜕𝑦′′And the boundary conditions,

𝜕𝐹


𝜕𝐹

𝜕𝑦′′𝛿𝑦′ −

𝑑

𝑑𝑥

𝜕𝐹


𝑥0

𝑥1

where

𝐹 𝑥, 𝑦 𝑥 , 𝑦′′ 𝑥 =𝑘

2𝑦′′ 𝑥

2− 𝑤 𝑥 𝑦(𝑥)

A rather tedious expansion of the variational derivative, using the explicit expression, immediately leads to the ODE,

𝑑4𝑦

𝑑𝑥4=𝑤 𝑥

𝑘.

Once this is satisfied, we only need to bother about the BCs. First recall that,

𝜕𝐹

𝜕𝑦′= 0 as, 𝐹 𝑥, 𝑦 𝑥 , 𝑦′′ 𝑥 depends explicitly only on 𝑦(𝑥)

and 𝑦′′ 𝑥 .


The boundary conditions are now given as

𝜕𝐹

𝜕𝑦′′𝛿𝑦′ −

𝑑

𝑑𝑥

𝜕𝐹


𝑥0

𝑥1

= 𝑘𝑦′′ 𝑥 𝛿𝑦′ − 𝑘𝑑

𝑑𝑥𝑦′′ 𝑥 𝛿𝑦

𝑥0

𝑥1

= 𝑘𝑦′′ 𝑥 𝛿𝑦′ − 𝑘𝜕

𝜕𝑥+ 𝑦′

𝜕

𝜕𝑦+ 𝑦′′

𝜕

𝜕𝑦′+ 𝑦′′′

𝜕

𝜕𝑦′′𝑦′′ 𝑥 𝛿𝑦

𝑥0

𝑥1

= 𝑘𝑦′′ 𝑥 𝛿𝑦′ − 𝑘 𝑦′′′ + 0 + 0 + 𝑦′′′ 𝛿𝑦 𝑥0

𝑥1

= 𝑘𝑦′′ 𝑥1 𝛿𝑦′ 𝑥1 − 2𝑘𝑦′′′ 𝑥1 𝛿𝑦 𝑥1 − 𝑘𝑦′′ 𝑥0 𝛿𝑦′ 𝑥0−2𝑘𝑦′′′ 𝑥0 𝛿𝑦 𝑥0

Let us now examine how these conditions are satisfied for the different loading conditions.

Before we proceed, first recall that a fourth-order ODE must have four boundary conditions to give a unique solution.


First recall that, 𝜕𝐹

𝜕𝑦′= 0 as, 𝐹 𝑥, 𝑦 𝑥 , 𝑦′′ 𝑥 depends explicitly only on 𝑦(𝑥)

and 𝑦′′ 𝑥 . The boundary conditions are now given as

𝜕𝐹

𝜕𝑦′′𝛿𝑦′ −

𝑑

𝑑𝑥

𝜕𝐹


𝑥0

𝑥1

= 𝑘𝑦′′ 𝑥 𝛿𝑦′ − 𝑘𝑑

𝑑𝑥𝑦′′ 𝑥 𝛿𝑦

𝑥0

𝑥1

= 𝑘𝑦′′ 𝑥 𝛿𝑦′ − 𝑘𝜕

𝜕𝑥+ 𝑦′

𝜕

𝜕𝑦+ 𝑦′′

𝜕

𝜕𝑦′+ 𝑦′′′

𝜕

𝜕𝑦′′𝑦′′ 𝑥 𝛿𝑦

𝑥0

𝑥1

= 𝑘𝑦′′ 𝑥 𝛿𝑦′ − 𝑘 𝑦′′′ + 0 + 0 + 𝑦′′′ 𝛿𝑦 𝑥0

𝑥1

= 𝑘𝑦′′ 𝑥1 𝛿𝑦′ 𝑥1 − 2𝑘𝑦′′′ 𝑥1 𝛿𝑦 𝑥1 − 𝑘𝑦′′ 𝑥0 𝛿𝑦′ 𝑥0+2𝑘𝑦′′′ 𝑥0 𝛿𝑦 𝑥0

Let us now examine how these conditions are satisfied for the different loading conditions.

Before we proceed, first recall that a fourth-order ODE must have four boundary conditions to give a unique solution.

Let us now examine how these conditions are satisfied for the different loading conditions. Before we proceed, first recall that a fourth-order ODE must have four boundary conditions to give a unique solution.

Case 1. Both ends fixed.Here, the displacements, 𝑦 𝑥0 as well as 𝑦 𝑥1 at both ends are given (zero for both in this case; what really matters is not that they are zero, but that they are prespecified). Furthermore, the slopes at the two ends, 𝑦′ 𝑥0 as well as 𝑦′ 𝑥1are also given (again, they are zero-prescribed is what matters).

Clearly, these conditions mean that

𝛿𝑦′ 𝑥1 = 0

𝛿𝑦 𝑥1 = 0

𝛿𝑦′ 𝑥0 = 0,

𝛿𝑦 𝑥0 = 0


Special Cases

It is essential that these conditions, prescriptions of no variation of the

function and its partial derivatives at the endpoints be satisfied a-priori

by any extremizing function to be admissible. They are called “Essential

Boundary conditions”. These are the conditions we need to be able to

get a unique solution to our ODE! With the essential boundary

conditions satisfied, it is clear that our extremizing conditions are now

satisfied:

𝑘𝑦′′ 𝑥1 𝛿𝑦′ 𝑥1 − 2𝑘𝑦′′′ 𝑥1 𝛿𝑦 𝑥1 − 𝑘𝑦′′ 𝑥0 𝛿𝑦′ 𝑥0 + 2𝑘𝑦′′′ 𝑥0 𝛿𝑦 𝑥0= 𝑘𝑦′′ 𝑥1 0 − 2𝑘𝑦′′′ 𝑥1 0 − 𝑘𝑦′′ 𝑥0 0 + 2𝑘𝑦′′′ 𝑥0 0 = 0.


Case 1 Continued

Clearly, these conditions mean that 𝛿𝑦′ 𝑥1 =?; 𝛿𝑦 𝑥1 =?; 𝛿𝑦′ 𝑥0 = 0; 𝛿𝑦 𝑥0 = 0

Only two of the parameters are prescribed. We cannot get a unique solution to our ODE now because we do not have a sufficient number of boundary conditions. The variation at the end 𝑥1 can no longer be assumed to vanish. The power of the variational method is that, in this case, it carries along with the extremization, the remaining boundary conditions needed to still obtain a unique solution. This conditions, occurring “naturally” with the first variation, are called “Natural Boundary Conditions”. In order to still obtain a vanishing of the boundary expression,

𝑘𝑦′′ 𝑥1 𝛿𝑦′ 𝑥1 − 2𝑘𝑦′′′ 𝑥1 𝛿𝑦 𝑥1 − 𝑘𝑦′′ 𝑥0 𝛿𝑦′ 𝑥0 + 2𝑘𝑦′′′ 𝑥0 𝛿𝑦 𝑥0when 𝛿𝑦 𝑥1 ≠ 0 and 𝛿𝑦′ 𝑥1 ≠ 0 we MUST have that,

𝑦′′ 𝑥1 = 0, 𝑎𝑛𝑑 𝑦′′′ 𝑥1 = 0.

Which in this case are proportional to the bending moment and shear forces at these points. The variational formulation yields these conditions naturally!


Case 2. One end (𝑥0), fixed, the other end free.

Here for each end, the displacement function is prescribed

𝛿𝑦′ 𝑥1 =?

𝛿𝑦 𝑥1 = 0

𝛿𝑦′ 𝑥0 =? ,

𝛿𝑦 𝑥0 = 0

(zero actually, but bear in mind, it is the prescription that matters), we still want the expression,

𝑘𝑦′′ 𝑥1 𝛿𝑦′ 𝑥1 − 2𝑘𝑦′′′ 𝑥1 𝛿𝑦 𝑥1 − 𝑘𝑦′′ 𝑥0 𝛿𝑦′ 𝑥0 + 2𝑘𝑦′′′ 𝑥0 𝛿𝑦 𝑥0

To vanish. Again, as an ODE problem, we need two more boundary conditions. At this point, it is easy to see that the Natural boundary conditions, following the previous development, are:

𝑦′′ 𝑥0 = 0 𝑎𝑛𝑑 𝑦′′ 𝑥1 = 0

Which has the physical interpretation that the shear forces vanish at these ends.


Case 3. Both ends simply supported

𝛿𝑦′ 𝑥1 =?

𝛿𝑦 𝑥1 = 0

𝛿𝑦′ 𝑥0 = 0,

𝛿𝑦 𝑥0 = 0

The remaining condition needed here is supplied by the

natural boundary condition, 𝑦′′ 𝑥1 .


Case 4. One end fixed,The other Simply Supported

For a homogeneous, isotropic elastic material undergoing small deformations given by the displacement vector 𝐮, assuming that 𝜆 and 𝜇 are the Lame’s coefficients, the energy functional,

𝐹 𝐮 =𝜇

2grad 𝐮 2 + 𝜆 + 𝜇 div2𝐮

can be minimized to find the equilibrium conditions. In Cartesian coordinates, we can write this functional in terms of the two multivariable functions by writing, 𝐮 = 𝑢 𝑥, 𝑦 𝐢 + 𝑣 𝑥, 𝑦 𝐣. Recall that

grad 𝐮 2 = tr gradT𝐮 grad 𝐮

=𝜕𝑢

𝜕𝑥

2

+𝜕𝑢

𝜕𝑦

2

+𝜕𝑣

𝜕𝑥

2

+𝜕𝑣

𝜕𝑦

2

and div 𝐮 =𝜕𝑢

𝜕𝑥+

𝜕𝑣

𝜕𝑦


Small Elastic Deformations

We therefore need to extremize the functional,

𝐽 𝑢, 𝑣 = Ω

𝜇

2grad 𝐮 2 + 𝜆 + 𝜇 div2𝐮 𝑑𝐴

= නΩ

𝜆

2+ 𝜇

𝜕𝑢

𝜕𝑥

2

+𝜕𝑣

𝜕𝑦

2

+𝜇

2

𝜕𝑢

𝜕𝑦

2

+𝜕𝑣

𝜕𝑥

2

+ 𝜆 + 𝜇𝜕𝑢

𝜕𝑥

𝜕𝑣

𝜕𝑦𝑑𝑥𝑑𝑦

Clearly,

𝐹 𝑢, 𝑣 = 𝐹 𝑢𝑥, 𝑢𝑦, 𝑣𝑥, 𝑣𝑦

Which corresponds to the case of multiple functions of the independent variable as we have in slide 45. There is no direct dependency on the functions 𝑢 and 𝑣 themselves. Note that,

𝜕𝐹(𝑢, 𝑣)

𝜕𝑢𝑥= 𝜆 + 2𝜇

𝜕𝑢

𝜕𝑥+ 𝜆 + 𝜇

𝜕𝑣

𝜕𝑦


Elastic Deformations


The Variational derivatives corresponding to this problem are,

𝜕𝐹

𝜕𝑢−

𝜕

𝜕𝑥

𝜕𝐹

𝜕𝑢𝑥−

𝜕

𝜕𝑦

𝜕𝐹

𝜕𝑢𝑦= 0

and 𝜕𝐹

𝜕𝑣−

𝜕

𝜕𝑥

𝜕𝐹

𝜕𝑣𝑥−

𝜕

𝜕𝑦

𝜕𝐹

𝜕𝑣𝑦= 0

After differentiating and ignoring the vanishing first term in each equation, we have,

𝜆 + 2𝜇𝜕2𝑢

𝜕𝑥2+ 𝜇

𝜕2𝑢

𝜕𝑦2+ 𝜆 + 𝜇

𝜕2𝑣

𝜕𝑦2= 0

and

𝜆 + 𝜇𝜕2𝑢

𝜕𝑥2+ 𝜇

𝜕2𝑣

𝜕𝑥2+ 𝜆 + 2𝜇

𝜕2𝑣

𝜕𝑦2= 0

Which are the Navier Equilibrium equations of elasticity. With boundary terms,

රΓ

𝑄 𝑥, 𝑦 𝛿𝑢 + 𝑅 𝑥, 𝑦 𝛿𝑣 𝑑𝑠

Where

𝑄 𝑥, 𝑦 =𝜕𝐹


𝜕𝐹

𝜕𝑢𝑦𝑛𝑦 and 𝑅 𝑥, 𝑦 =

𝜕𝐹


𝜕𝐹

𝜕𝑣𝑦𝑛𝑦

MCE 511 Finite Element Analysis - OA...

Documents

Transcript of MCE 511 Finite Element Analysis - OA...